Let $\mathbf{Q}$ be the group of rational numbers under addition and let $\mathbf{Q}^\times$ be the group of nonzero rational numbers under multiplication. Find the order of each element in $\mathbf{Q}$ and $\mathbf{Q}^\times$.

I know the order of an element is the least positive integer, $n$, such that $a^n = e$.
$\mathbf Q= \{0,1,\displaystyle\frac{1}{2}, \frac{1}{3}, \frac{1}{4} \dots\}$ and
$\mathbf{Q}^\times = \{1, \displaystyle\frac{1}{2}, \frac{1}{3}, \frac{1}{4} \dots\}$

The identity under function addition is such that you get $0$ and the identity under multiplication is such that you get $1$.

How would I go about finding the order of the elements in either group?

2 Answers
2

Remember, $a^n$ in this context doesn't (necessarily) refer to $a$ to the power of $n$, in the traditional sense. What it means is $a * a * a * \ldots * a$, $n$ times, where $*$ is whatever the group operations is. In the case where $*$ is multiplication, then this becomes $a^n$ in the traditional sense. When $*$ is addition, this refers to $a + a + a + \ldots + a = na$.

So, if we take an arbitrary rational number $q \in \mathbb{Q}$, then what's its order in the group under addition? We are solving for $nq = 0$ (as $0$ is the additive identity) for smallest $n \ge 1$.

Most of the time, this has no solution. In particular, when $q \neq 0$, then $n = 0$ is the only solution. Since there is no positive integer $n$ such that $nq = 0$, this means $q$ has infinite order.

On the other hand, when $q = 0$, then any $n$ will do, so the smallest $n \ge 1$ is $n = 1$. Thus, the order is $1$ (as it always is for the identity).

What about for $\mathbb{Q}^*$? I'll leave it to you. I'll give you a hint: there are exactly two elements of finite order.

As you stated, the order of an element $a$ is the minimum $n\in \mathbf{N}_{>0}$ such that $a^n=e$ (or $na=0$ in additive notation). If no such $n$ exists, then we say that the order of the element is infinite.

Now, let $x\in \mathbf{Q}$. If $x=0$, well, $1\cdot0=0$. If $x\ne 0$, can we find and $n$ sufficiently large that $nx=0$?

Similarly, if $y\in \mathbf{Q}^\times$, there are three cases. If $y=1$, then $y^1=1$. If $y=-1$, then $y^2=1$. If $y\ne \pm1$, can we find a finite $n$ such that $y^n=1?$